Tabulation Bottom Up DP in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array with zeros.

DSA C#

int[] dp = new int[n + 1];
for (int i = 0; i <= n; i++) {
    dp[i] = [1];
}

Drag options to blanks, or click blank then click option'

Ai

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 1 causes incorrect base values.

Using -1 may cause wrong results in summation problems.

The DP array should be initialized with zeros to represent base cases before filling in values.

2fill in blank

medium

Complete the code to fill the DP array using the bottom-up approach.

DSA C#

for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] + [1];
}

Drag options to blanks, or click blank then click option'

Ai

Bdp[i]

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Adding i instead of 1 changes the count incorrectly.

Adding dp[i] causes self-reference and errors.

We add 1 to the previous dp value to count ways or steps incrementally.

3fill in blank

hard

Fix the error in the DP update step to avoid index out of range.

DSA C#

for (int i = 1; i <= n; i++) {
    if (i - 2 >= 0) {
        [1]
    }
}

Drag options to blanks, or click blank then click option'

Adp[i] = dp[i] + dp[i - 1];

Bdp[i] = dp[i - 1] + dp[i - 2];

Cdp[i] = dp[i - 2];

Ddp[i] = dp[i - 1];

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp[i] + dp[i-1] causes double counting.

Assigning dp[i] = dp[i-2] loses dp[i-1] contribution.

The correct update sums dp[i-1] and dp[i-2] to build the current dp[i] value.

4fill in blank

hard

Fill both blanks to complete the nested loop for a 2D DP table.

DSA C#

for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= [1]; j++) {
        dp[i, j] = dp[i - 1, j] + dp[i, [2]];
    }
}

Drag options to blanks, or click blank then click option'

An

Bj - 1

Cm

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using m instead of n for inner loop limit.

Using i-1 instead of j-1 for dp index.

The inner loop runs up to n, and dp[i, j] depends on dp[i, j-1].

5fill in blank

hard

Fill all three blanks to complete the DP dictionary comprehension equivalent in C#.

DSA C#

var result = new Dictionary<string, int>();
foreach (var word in words) {
    result[[1]] = word.[2];
    if (result[[3]] == 0) {
        result[[3]] = 1;
    } else {
        result[[3]] += 1;
    }
}

Drag options to blanks, or click blank then click option'

Aword

BLength

DCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using Count instead of Length for string length.

Using inconsistent keys for dictionary access.

We use the word as key, Length property for length, and word again as key to update count.

Practice

1. What is the main idea behind Tabulation Bottom Up DP in dynamic programming?

easy

A. Start solving from the main problem and break it down recursively

B. Solve the problem by recursion without storing results

C. Use random guesses to find the solution

D. Build a table from the smallest subproblems up to the main problem

Solution

1. Step 1: Understand bottom-up approach

   Tabulation means filling a table starting from the smallest subproblems, building up to the final answer.

2. Step 2: Compare with other methods

   Unlike recursion without memoization, tabulation avoids repeated work by storing results in a table.

3. Final Answer:

   Build a table from the smallest subproblems up to the main problem -> Option D

4. Quick Check:

   Tabulation = Bottom-up table filling [OK]

Hint: Think: solve small parts first, then bigger [OK]

Common Mistakes:

• Confusing tabulation with recursion
• Ignoring the order of filling the table
• Not storing intermediate results

2. Which of the following is the correct way to initialize a tabulation array for Fibonacci numbers in C#?

easy

A. int[] dp = new int[n + 1]; dp[0] = 1; dp[1] = 0;

B. int[] dp = new int[n]; dp[0] = 1; dp[1] = 1;

C. int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1;

D. int[] dp = new int[n]; dp[0] = 0; dp[1] = 1;

Solution

1. Step 1: Check array size and base cases

   For Fibonacci, dp array size should be n+1 to include dp[n]. Base cases are dp[0]=0 and dp[1]=1.

2. Step 2: Verify initialization correctness

   int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; correctly initializes dp with size n+1 and sets dp[0]=0, dp[1]=1.

3. Final Answer:

   int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; -> Option C

4. Quick Check:

   Correct array size and base values [OK]

Hint: Remember Fibonacci base: 0 and 1 at dp[0], dp[1] [OK]

Common Mistakes:

• Using wrong array size (n instead of n+1)
• Swapping base values dp[0] and dp[1]
• Not initializing base cases

3. What will be the output of the following C# code using bottom-up DP to find the number of ways to climb 3 steps if you can take 1 or 2 steps at a time?

int n = 3;
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
}
Console.WriteLine(dp[n]);

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Understand dp array meaning

   dp[i] stores number of ways to reach step i. Base: dp[0]=1 (one way to stand still), dp[1]=1 (one way to step 1).

2. Step 2: Calculate dp values up to n=3

   dp[2] = dp[1] + dp[0] = 1 + 1 = 2; dp[3] = dp[2] + dp[1] = 2 + 1 = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Ways to climb 3 steps = 3 [OK]

Hint: Sum ways to previous two steps for each step [OK]

Common Mistakes:

• Confusing dp indices or base cases
• Off-by-one errors in loop
• Misunderstanding dp meaning

4. Identify the error in this bottom-up DP code snippet for calculating factorial of n:

int n = 5;
int[] dp = new int[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1] * i;
}
Console.WriteLine(dp[n]);

medium

A. No error, code works correctly

B. Array dp should be of size n, not n+1

C. dp[0] should be initialized to 0

D. Loop should start from i = 0

Solution

1. Step 1: Check array size and initialization

   dp size n+1 is correct to store factorials from 0 to n. dp[0] = 1 is correct as 0! = 1.

2. Step 2: Verify loop and calculation

   Loop from i=1 to n correctly calculates dp[i] = dp[i-1] * i, which is factorial definition.

3. Final Answer:

   No error, code works correctly -> Option A

4. Quick Check:

   Factorial DP code is correct [OK]

Hint: Check base case and loop range carefully [OK]

Common Mistakes:

• Wrong dp array size causing index errors
• Incorrect base case initialization
• Starting loop at wrong index

5. Given a list of coin denominations coins = {1, 2, 5} and a target amount amount = 5, which bottom-up DP approach correctly computes the minimum number of coins needed to make the amount?

int amount = 5;
int[] dp = new int[amount + 1];
for (int i = 1; i <= amount; i++) dp[i] = int.MaxValue;
int[] coins = {1, 2, 5};
for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin <= i && dp[i - coin] != int.MaxValue) {
            dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
        }
    }
}
Console.WriteLine(dp[amount]);

hard

A. 1

B. 2

C. 3

D. 5

Solution

1. Step 1: Understand dp meaning and initialization

   dp[i] stores minimum coins to make amount i. Initialized dp[0]=0 (default), others to MaxValue.

2. Step 2: Calculate dp values for amount=5

   Calculate dp[1]=1 (coin 1), dp[2]=1 (coin 2), dp[3]=2 (1+2), dp[4]=2 (2+2), dp[5]=1 (coin 5). Minimum coins for 5 is 1.

3. Step 3: Re-check code logic

   Code correctly updates dp[i] with minimum coins. Output dp[5] = 1.

4. Final Answer:

   1 -> Option A

5. Quick Check:

   Minimum coins for 5 = 1 (coin 5) [OK]

Hint: Check dp updates with all coins for each amount [OK]

Common Mistakes:

• Not initializing dp array properly
• Ignoring dp[i - coin] == int.MaxValue check
• Confusing minimum with maximum coin count